Produce the number 2014 without any numbers in your source code

Question

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It's 2017 2018 2019 2020 2021 2022 2023 already, folks, go home.

So, now that it's 2014, it's time for a code question involving the number 2014.

Your task is to make a program that prints the number 2014, without using any of the characters 0123456789 in your code, and independently of any external variables such as the date or time or a random seed.

The shortest code (counting in bytes) to do so in any language in which numbers are valid tokens wins.

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Leaderboard:

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Answer 1

J (13)

#.a.i.'_!!! '

Interprets the ASCII value of _!!! (95 33 33 33 32) as a binary number (it's weird that this is possible, I agree). This produces 2014.

J (15)

This one doesn't use any character strings. It's based on the weird coincidence that the sum of the first 46 primes is 4028: double 2014.

-:+/p:i.<:+:_bn

If anyone knows of a shorter way than <:+:_bn to represent 45 (preferably without strings), please let me know.

Answer 2

C#, 4 characters, 5 bytes

+'ߞ'

Note: you need LINQPad to run it, not Visual Studio. LinqPad is good for CodeGolfing in C#.

Answer 3

JavaScript, 24 bytes

A bit long, but no idea how this way got left out...

alert("ߞ".charCodeAt())

Explanation

The character ߞ is obtained by doing String.fromCharCode(2014) . Thus the code is actually just converting that character back to its character code and alerting it.

Thanks to hsl for this shorter version

Answer 4

Python 2 (19 bytes, ASCII only, CPython-specific)

print hash("w_'qe")

Tested only on 64-bit, but I assume/hope that since 2014 is small and positive the results would be the same on 32-bit? Originally tested on Python 3, but ProgramFOX confirms it also works on Python 2.

Python 3 (31 bytes, ASCII only)

print(ord("\N{NKO LETTER KA}"))

Quite fond of this one, even though better solutions exist. The equivalent Python 2 code is no shorter, as it required a u string prefix.

Answer 5

Insomnia, 7

Each line is one program doing the same thing: print 2014 to output stream.

e}u#Hi-
e}u#Hs-
e}u#H}-
e}g#*i-
e}g#*s-
e}g#*}-
e}gKHi-
e}gKH}-
e}gKxi-
e}gKxs-
e}gKx}-
e}u#dK-
e}u#eK-
e}u#fK-
e}gKdK-
e}gKeK-
e}gKfK-

Answer 6

CMD - 42 bytes

set/aa=f
set/a%a%xAAA-%a%xFF-%a%xFF-%a%xCE

The 'trick' is that when using the /a switch on the set command, letters (and other invalid characters) are evaluated as 0. I then just use hexadecimal to evaluate 2014. The 0 is needed because in CMD hexadecimal must be expressed with the leading 0x. There is almost definitely a shorter way to get to 2014...

%a%xAAA - %a%xFF - %a%xFF - %a%xCE = 2730 - 255 - 255 - 206 = 2014

Answer 7

Python (30 chars)

(10 + 9) * 106 = 2014

(ord('\t')+ord('\n'))*ord('j')

Answer 8

Hassium, 77 Bytes

Really excited about this one. It gets Math.pi and divides it by itself in variable a (1), then uses increment and basic math operators to get it to 2014.

use Math;func main(){a=Math.pi;a/=a;print(++a)print(a-a)print(a/a)print(a*a)}

Run online and see expanded here

Answer 9

Jolf, 3 bytes

(It's almost 2016. Language obviously postdates question.)

@ߞ
@  Get charcode of next character
 ߞ charcode 2014

Answer 10

JavaScript, 9 bytes

btoa`ÛMx`

This Base-64 encodes ÛMx to make 2014.

Answer 11

Lua, 32 bytes

print(#'XX'..#''..#'X'..#'XXXX')

This makes use of the length operator # used on string literals to get numbers, and the lengths are concatenated (with ..) to print 2014

Answer 12

Brain-Flak, 50 bytes

((((((()()()()()){}){})){}{}()){()()({}[()])}{}())

Try it online!

Brain-flak is great for restricted source challenges because there are only 8 valid character to begin with: brackets. (e.g. []{}()<>).

This was made possible with the help of @ASCII-only's integer metagolfer, which is currently hosted online at brain-flak.github.io/integer

Answer 13

Brain-Flak, 46 + 3 = 49

Try it online

(([()]([(()()())]((((({}){}){}()){}){})))()())

This one uses the -A flag for +3 bytes. It pushes the ASCII values for 2014 to the stack which outputs as 2014 in ASCII mode.

Answer 14

Sinclair ZX81 15 bytes 10 bytes

 PRINT CODE "=";CODE ":"

As the ZX81 has a non-ASCII compatible character set, the character code for = is 20 and for : it is 14 - simples.

Answer 15

Fourier, 12 bytes

Non-competing: Fourier is newer than the challenge

I know I'm two years too late, but it seemed like a fun challenge to do in Fourier.

^^ovvo^o^^^o

The command ^ increments the value of the accumulator (which starts at zero) and v decrements the value of the accumulator.

Try it online!

Answer 16

Lean Mean Bean Machine, 53 bytes

\\\
O))\ \
#)) o )
 ))u ))
 ))  ))
 ))  )u
 ))  ~
~~~

I like this.

Answer 17

Jelly, 3 bytes

⁽¥Æ

Try it online!

Answer 18

J, 21 bytes

,":,.$,:~}.,:,:'golf'

Try it online!

               'golf'  One dimensional array
           ,:,:        Itemize twice (1x1x4 array)
         }.            Drop first element (0x1x4)
      ,:~              Append to itself as distinct items (2x0x1x4)
     $                 Get dimensions (2 0 1 4)
   ,.                  Flatten items, essentially prints 2014 vertically.
                       (so there are no spaces)
,":                    To strings, flatten.    

20 bytes

#.(#_),,~(,~,~#_),%_
#.(#_),,~(,~$,._),%_

15 bytes

do'bbbc',~":_bd

11 bytes

,":_bk,:_be

Answer 19

><>, 7 bytes

aa+nen;

Try it online!

aa                      +                           n                                    e                 n                                    ;
^^                      ^                           ^                                    ^                 ^                                    ^
Push 10 to stack twice, add top two items in stack, print top item in stack as a number, push 14 to stack, print top item in stack as a number, stop.

Answer 20

JavaScript, various solutions: 131, 44, 43, 30, 28, and 16 characters (16 bytes)

Original answer:

JS-fuck inspired JavaScript, 131 characters:

(!![]<<(!![]<<!![])<<!![]<<(!![]<<!![])<<!![]<<(!![]<<!![])<<!![]<<(!![]<<!![]))-((!![]<<!![])<<!![]<<(!![]<<!![])<<!![])-!![]-!![]

Shifty truth edition:

(true<<(true<<true)<<true<<(true<<true)<<true<<(true<<true)<<true<<(true<<true))-((true<<true)<<true<<(true<<true)<<true)-true-true

Addenda

Edit: I'm having too much fun with this... combining type coercion with hex strings, 44 characters.

a=+[];(a+'xade')-(a+'x'+(+!a+!a+!a)+a+a)
a=+[];(a+'xade')-(a+'x'+(++a+a+a)+--a+a)

a is 0, 0xade is 2782. We need to subtract 768 to get 2014. 768 is 0x300. (a+'x'+(+!a+!a+!a)+a+a) and (a+'x'+(++a+a+a)+--a+a) are two ways of producing "0x300", so the final result is "0xade" - "0x300", which JavaScript coerces back to numbers, resulting in 2014.

Here is a radix 36-based parseInt solution, 43 characters:

(a=+[]);parseInt(++a+'jy',a+++a+++''+(a+a))

First, we initiate a variable a at 0 (clearly this requires non-strict mode), increment it to 1, concatenate with 'jy' - 1jy happens to be 2014 in radix 36. The quickest way call parseInt at this radix is to generate the string "36" and abuse type coercion again: a++ + a++ results in 3, with a set to 3 as well, which means (a+a) is 6, so a+++a+++''+(a+a) results in "36", resulting in 2014.

After coming up with these two solutions I started looking at other JavaScript answers (thanks for the tip, Scrooble!), to see if combining ideas from other people gives interesting results.

Zaq's approach can be shortened with the hex string trick to 28 characters:

-~[]+!![]+[+[]]+ +(+[]+'xe')

How it works: -~[]+!![]+[+[]] results in "20", +[]+'xe' results in "0xe". +"0xe" is 14, so "20" + +"0xe" becomes "20" + 14, which becomes "2014".

Now here's a fun fact: 20 in hexadecimal is 0x14! Oh, and 14 is 0xe. So +('0x' + 0xe) produces 20. Combining this with the above results in a 30 character solution:

_=+[]+'x';+(_+ +(_+='e'))+[+_]"

Finally, initially inspired by Dan Prince's answer I came up with what might be the shortest possible JavaScript solution abusing hex-strings, at 16 characters:

+[]+'xfbc'>>!![]

0xfbc is 4028. Shift right by one, and we have 2014. If anyone knows of a shorter way to generate 0 or 1, let me know.

Edit2: having exhausted the hexadecimals, we should of course also look at binary and octal notation!

// +'0b11111011110'
+(a=+[],a+++'b'+(a+(((a+=[a]+a+a+(+[])))+a)))
// +'0o3736'
+(a=-~[],a=a+++a+++''+(a+a),+[]+'o'+((+a+1)+a))

Answer 21

Pyt, 27 bytes

ɳąḞḞ⬠⬠⬠π⎶⁻⦋ĐąžΠ²+ĐŚřƩ½*-⁻⁻Ɩ

Try it online!

Not exactly a serious contender, just had some fun.

ɳ             push '0123456789' as string
 ą             convert to array of digits [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
  ḞḞ            for each item in array, replace with xth fibonacci # (2x) [1, 1, 2, 3, 8, 34, 377, 17711, 9227465, 225851433717]
   ⬠⬠⬠          for each item in array, replace with xth pentagonal # (3x) [1, 1, 1820, 66045, 240027425, 29321506727800, 6947548864499411875070L, 165405818231059923692911546880492501L, 898044801648686628863443901192030771814779461710865094720L, 115670237695821250427139838385782853032222541808893547195455834936957002151009052998969975100L]
       π⎶⁻           push pi, round, and decrement (2)
          ⦋         get the 2th element of that list (1820)
           Đ         Duplicate 1820
            ąž        make an array of digits and remove zeroes [1, 8, 2]
              Π        multiply together (16)
               ²+       square and add (2076)
                 ĐŚ      Duplicate and sum digits (15)
                   řƩ     make a range from 1 to 15 and sum all (120)
                     ½*    multiply by 1/2 (60)
                       -    subtract  (2016.0)
                        ⁻⁻   decrement (2x)   (2014.0)
                          Ɩ   cast to integer (2014)
implicit output

Answer 22

JavaScript (ES6), 52 46 bytes

t=-~'';console.log(`${++t}${t-t}${t/t}${t*t}`)

Try it online!

Will log through console.log the number 2014 as a string.

Thanks to Jacob for multiple optimizations saving 6 bytes

Explanation

t=-~'';

In JavaScript, ~~ will convert the proceeding value to a number, in this case, true equates to 1.

Set the value of t to 1 by using ~ on an empty string, which would equate to -1, then take the opposite of that number, 1.

For more info about tilde in JavaScript, see this article.

console.log(` ... `)

Logs the template string ... with ${} expressions available, where ... includes:

${++t}

Sets w to t+t, which would be 2, which would return the number 2. Added to string.

Set t to itself + 1, and display the final result, 2

${t-t}

Displays t-t, which would be 0, which would return the number 0.

${t/t}

Takes the value of t and divides by itself, returning 1.

${t*t}

Takes the value of w*w*w*w-w (or w ^ 4 - 1), where w (as previously set) is 2, and subtracts w from it, and returns the result. Added to string.

Takes the value of t*t (or t ^ 2), where t (as previously set) is 2.

The added expressions equate to 2014, which ... is in the log.

Answer 23

Lost, 57 23 22 bytes

<^/*(
 )/@+
">>v
^?%<<

My first Lost answer. Thought I'd start with an easy one.

Byte-count more than halved (-35 bytes) thanks to @JoKing.

Try it online or verify that it's deterministic.

General explanation about Lost:

Let me start with an explanation of Lost itself. Lost is a 2D path-walking language. Most 2D path-walking languages start at the top-left position and travel towards the right by default. Lost is unique however, in that both the start position AND starting direction it travels in is completely random. So making the program deterministic, meaning it will have the same output regardless of where it starts or travels, can be quite tricky.

A Lost program of 2 rows and 5 characters per row can have 40 possible program flows. It can start on any one of the 10 characters in the program, and it can start traveling up, down, left, or right.

In Lost you therefore want to lead everything to a starting position so it follows the designed path you want it to. In addition, you'll usually have to clean up the stack when it starts somewhere in the middle.

Program explanation:

The 22-bytes program is similar as the previous 23-bytes program below, but with a smarter path to save that byte:

v<<<<<>>>>>
>%?"^ <"*+@

Let me start with an explanation of the 23-bytes program:

The "^ <" will push the character-codepoints for the three characters in the string, being 94 32 60 respectively. The * multiplies the top two, and + adds the top two of the stack, so it becomes 94+(32*60), which results in 2014.

The @ will terminate the program, but only if the safety is 'off'. When the program starts the safety is always 'on', otherwise the program starting at the exit character immediately terminates without doing anything.
The % will turn the safety 'off'. So as soon as the % is encountered and the safety is 'off', the program can be terminated with an @.

The ? is to clean up the stack if it started somewhere in the middle.

And finally the v<<<<<>>>>>, > and use of ^ < in the string are to lead the program path towards the correct starting position for it to correctly print 2014. Note that the top line could have been v<<<<<<<<<<, but that the reversed part >>>>> will wrap-around to the other side, making the path shorter and therefore the performance slightly better. The byte-count remains the same anyway, so why not.

---

Now for the 22-bytes solution, and how it actually is the same as the 23-bytes solution, but with a different path.

The arrows are still used to lead the path into the given direction. The / are used as a mirror. So if we go from right to left and encounter the /, it will continue downwards; if we go from the top to the bottom and encounter the /, it will continue towards the left; etc.

The ( will pop the top value on the stack and push to to the scope, and the ) will do the reversed: it pops from the scope, and pushes it back to the stack.

So regardless of where we start and in which direction we travel, the path leads towards the first < of the bottom row. From there, the program flow travels in this order:

%?^        Direction changed upwards
" <^" <    Direction changed towards the left
(*/        Direction changed downwards
/          Direction changed towards the left
) +@

So it will:

• Turn the safety 'off' with %;
• Clean the stack with ?;
• Push the character-codepoints for " <^", which are 32 60 94 respectively;
• Pop the 94 and store it in the scope with (;
• Multiply the 32 60 with *, resulting in 1920;
• Push the 94 from the scope back onto the stack with );
• Add the 1920 94 together with +, resulting in 2014;
• And then terminates the program with @, implicitly outputting the top of the stack.

Answer 24

bc, 7 bytes. Try it Online!!

K*ZZ+Y

bc, 8 bytes. Try it Online!!

K*A*A+E

Which needs 14 bytes to run in bash:

bc<<<"K*A*A+E"

In bc the upper (single) letters maintan their meaning as a number in 10-36 range in any input base.

A previous approach changed the input base:

echo ibase=D\;BBC|bc

Make numeric base 13 (D) and print BBC in that base --> 2014.

Answer 25

Flobnar, 17 14 bytes

<+\@:!
+<>.!..

Try it online!

Explanation:

   @       Start, going left

  \        Push to the stack the value from the bottom row
  >

   .!..   Several print statements we will get back to
<   :!    Add the not of the top of the stack to itself
+         This is !0+!0 = 2

      .   Print the 2

     .    Print the result of the print (0)

   .!     Print the result of the not of the print (1)
  \       Continue forward after pushing the zero to the stack

<+   :!   Add the same 2 from the beginning to itself
+<        (!0 + !0)+(!0 + !0) = 4
          And print implicitly as the last value

Answer 26

05AB1E, 22 10 8 7 6 bytes

T·žvÍJ

Try it online!

How it works:

• T pushes 10, · doubles it, we get 20.
• žv pushes 16, Í subtracts 2, we get 14.

Lastly, we concatenate them using J!

Answer 27

C#, 25 bytes (24 characters)

Console.Write((int)'ߞ');

Try it online!

Explanation

The decimal Unicode of ߞ is 2014, so you can just cast it to an int and 2014 is printed.

Answer 28

Python, 52 49 chars

from math import*
print(int((e*pi+e)**pi+e/e+e))

Works in Python 2 and Python 3.

Answer 29

Labyrinth, 12 bytes

))!!)!))))!@

Try it online!

))))_#!!#!!@

Try it online!

))_#"
 ) !!@

Try it online!

))!"
)
)#!!@

Try it online!

I tried both linear and complex layouts, but I can't figure out how to remove a single byte from any of these programs.

Answer 30

Straw, 13 bytes

(…………………σ)«$>

« sum the codepoint of all characters in a string, $ convert from unary to decimal and > is the print operator.

Try it online